The polynomial

Solution:

Given, $\quad p(x)=x^{4}-2 x^{3}+3 x^{2}-a x+3 a-7$

When we divide $p(x)$ by $x+1$, then we get the remainder $p(-1)$.

Now, $\quad p(-1)=(-1)^{4}-2(-1)^{3}+3(-1)^{2}-a(-1)+3 a-7$

$=1+2+3+a+3 a-7=4 a-1$

According to the question, $p(-1)=19$

$\Rightarrow \quad 4 a-1=19$

$\Rightarrow \quad 4 a=20$

$\therefore \quad a=5$

$\therefore$ Required polynomial $=x^{4}-2 x^{3}+3 x^{2}-5 x+3(5)-7$ [put $a=5$ in $p(x)$ ]

$=x^{4}-2 x^{3}+3 x^{2}-5 x+15-7$

$=x^{4}-2 x^{3}+3 x^{2}-5 x+8$

When we divide $p(x)$ by $x+2$, then we get the remainder $p(-2)$.

Now, $\quad p(-2)=(-2)^{4}-2(-2)^{3}+3(-2)^{2}-5(-2)+8$

$=16+16+12+10+8=62$

Hence, the value of $a$ is 5 and remainder is 62 .